aa r X i v : . [ m a t h . C A ] M a y From Symmetry to Monotonicity
Bernd Kawohl and David Krejˇciˇr´ık
12 March 2018
Symmetry is a fascinating notion in science, primarily perhaps due to the fact thatthe symmetry of geometric objects evokes the aesthetic and elegant perception of everyhuman being. More fundamentally, the symmetries of spacetime are closely related toconservation laws in physics. In this note we show how one can use a symmetry of atwo-dimensional surface to prove a monotonicity property of a one-dimensional function.A couple of years ago Jan Ubøe gave an unconventional proof of the fact that a certainmapping f ( x ) which depends on a positive parameter a is increasing in x if a > a <
1, see [1]. In fact, f is given by f ( x ) = φ ( ax ) − φ ( x ) x (cid:0) Φ( ax ) + Φ( x ) (cid:1) where φ ( x ) = e − x is the derivative of Φ( x ) = R x φ ( u ) du . This behaviour of f is usedto describe certain effects in a so-called newsvendor model in operations research [2].The author of [1] reckons that Euler would have liked his proof (perhaps because oftransferring the problem to cumbersome tasks involving infinite sums), but also that avery short proof is lurking around somewhere . We believe that our alternative proof offerssimpler geometric arguments and is more tractable and appealing.The proof that we present makes repeated use of the facts that:(a) φ can be extended as an even function on R ;(b) the derivative satisfies a recurrent formula, namely φ ′ ( x ) = − xφ ( x ).First we observe that the sign of f ′ ( x ) is determined by the sign of h ( x ) = (cid:0) axφ ′ ( ax ) − xφ ′ ( x ) (cid:1)(cid:0) Φ( ax ) + Φ( x ) (cid:1) − (cid:0) φ ( ax ) − φ ( x ) (cid:1)(cid:0) Φ( ax ) + Φ( x ) (cid:1) − (cid:0) φ ( ax ) − φ ( x ) (cid:1)(cid:0) axφ ( ax ) + xφ ( x ) (cid:1) . (1)Then we rewrite each of the sums of two terms as integrals with help of properties (a)and (b) above and arrive at a double-integral representation h ( x ) = Z ax − x Z ax − x φ ( u ) φ ( v ) Γ( u, v ) dudv (2)with Γ( u, v ) = 2( u + v )( u − v ) . v = − u and positive above it. It is due to the obvious symmetry of φ stated in (a) abovetogether with a hidden symmetry encoded in the convolution kernelΓ( u, v ) = − Γ( − u, − v )valid for all u, v ∈ R . Consequently, since we are integrating over the rectangle ( − x, ax ) × ( − x, ax ), the odd symmetry immediately implies that ( cf. Figure 1): • for a = 1 the domains of positivity and negativity of Γ are equilibrated, so h ( x )equals zero (and f is constant, in fact it is identically equal to zero); • for a > h ( x ) ispositive (and f is strictly increasing); • for a < h ( x ) isnegative (and f is strictly decreasing).In summary, we have obtained the desired monotonicity of the one-dimensional func-tion f as a consequence of the symmetry properties of the two-dimensional surface Γ. - - - - uv - - Figure 1: Graph of the surface Γ and its contour plot. Integration over the green square(corresponding to x = 1 and a = 1 in the picture) gives zero because of the odd symmetryof Γ with respect to the line u = − v , while the integral over the rectangle expanded bythe lighter green colour into the positive (brownish) region of Γ (corresponding to a = 1 . axφ ′ ( ax ) − xφ ′ ( x ) = − (cid:0) a x φ ( ax ) − x φ ( x ) (cid:1) = − Z axx (cid:0) u φ ( u ) (cid:1) ′ du = − Z axx uφ ( u ) du + 4 Z axx u φ ( u ) du = − Z ax − x uφ ( u ) du + 4 Z ax − x u φ ( u ) du.
2n the last equality we have used the fact that uφ ( u ) and u φ ( u ) are odd functions of u .Similarly we writeΦ( ax ) + Φ( x ) = Z ax − x φ ( u ) du,φ ( ax ) − φ ( x ) = Z ax − x φ ′ ( u ) du = − Z ax − x uφ ( u ) du,axφ ( ax ) + xφ ( x ) = Z ax − x ( uφ ( u )) ′ du = Z ax − x φ ( u ) du − Z ax − x u φ ( u ) du. Putting these formulae together one can rewrite h as follows: h ( x ) = 4 (cid:18)Z ax − x u φ ( u ) du (cid:19) (cid:18)Z ax − x φ ( u ) du (cid:19) − (cid:18)Z ax − x uφ ( u ) du (cid:19) (cid:18)Z ax − x u φ ( u ) du (cid:19) . Now the main idea is to rewrite the product of integrals ( R g ( u ) du )( R j ( v ) dv ) as adouble integral RR g ( u ) j ( v ) dudv and realise that h ( x ) = Z ax − x Z ax − x φ ( u ) φ ( v ) ˜Γ( u, v ) dudv = Z ax − x Z ax − x φ ( u ) φ ( v ) ˜Γ( v, u ) dudv, where ˜Γ( u, v ) = 4( u − uv ). Here, the second equality is due to the fact that the role ofvariables u and v can be interchanged. Therefore, ˜Γ( u, v ) can be replaced by ˜Γ( v, u ) andalso by the arithmetic meanΓ( u, v ) = ˜Γ( u, v ) + ˜Γ( v, u )2 = 2( u + v )( u − v ) , so that (2) holds. Acknowledgment
D.K. was partially supported by the GACR grant No. 18-08835S and by FCT (Portugal)through project PTDC/MAT-CAL/4334/2014.
References [1] J. Ubøe, A heroic proof,
Math. Intell. (2015) pp. 72–74.[2] J. Ubøe, J. Andersson, K. J¨ornsten, J. Lillestøl, & L. Sandal, Statistical testing ofbounded rationality with applications to the newsvendor model, European J. Oper.Res. (2017) pp. 251–261.